Mathematics - 2023-06-06 (page 1 of 2)

shintuku

00:21

Let $K[x,y]_{\langle x,y\rangle}$ be the localization of $K[x,y]$ at the maximal ideal $\langle x,y \rangle$. Is $K[x,y]_{\langle x,y\rangle}[1/y]$ a local ring?

Ted Shifrin

Gee. I wonder why it’s called localization.

shintuku

is it obvious to you it is local or nonlocal?

Ted Shifrin

Have you done simple examples, like not with polynomials?

shintuku

i'm studying valuation rings and that's the simplest example i could think of of extending a ring that is local but not a valuation ring

Ted Shifrin

You’re in over your head. Start with $\Bbb Z$.

shintuku

00:29

problem is all localization at prime ideals there are valuation rings

i want the behaviour of extending local, nonvaluation rings

am slowly working through the $K[x,y]_{\langle x,y\rangle}[1/y]$ example step by step

Ted Shifrin

So why is $\Bbb Z_{(5)}$ local?

shintuku

i lied

$\mathbb Z_{(5)}$ is local because any element of $\mathbb Z_{(5)}$ who has a numerator divisible by $5$ is a nonunit

and that set is an ideal

Ted Shifrin

So is there an obvious maximal ideal?

shintuku

yes: any nonunit is contained in this single ideal

if you extend then this ideal by any other element not already in it, we get the entire ring

Ted Shifrin

Is there a good way to describe it more succinctly?

shintuku

00:40

let me think

$(\mathbb Z - \langle 5\rangle)^{-1}\langle 5\rangle$?

Ted Shifrin

Yeah, just the ideal generated by 5.

shintuku

the issue is, in my above mentioned example the extension doesn't make it a localization anymore and we're making elements who were previously nonunits into units

Ted Shifrin

Huh? I have no idea what you’re talking about.

shintuku

the example $K[x,y]_{\langle x,y\rangle}[1/y]$

Ted Shifrin

Oh, I missed the 1/y.

shintuku

00:47

previously, an element of $K[x,y]$ who's numerator was in $\langle y \rangle$ was a nonunit, now it becomes a unit

Ted Shifrin

So how is that related to localizing just at $(x)$?

shintuku

it's not a localization anymore, or at least, not a localization given by the natural homomorphism for localization at a prime ideal

Ted Shifrin

So answer my question.

shintuku

i'll think about it 2 seconds

i'm not sure, i need to think on it a while

shintuku

01:24

i have the $K[x,y]_{(x,y)}[1/y] \subseteq K[x,y]_{(x)}$ direction, I'm trying to get the other one

shintuku

01:57

any tips?

shintuku

02:47

@TedShifrin Notice $\frac{1}{x+y} \in K[x,y]_{(x)}$. Suppose we have $\frac{1}{x+y} \in K[x,y]_{(x,y)}[1/y]$. Then $\frac{1}{x+y} = \frac{p_0}{p'_0} + \frac{p_1}{p'_1y}+\cdots+\frac{p_n}{p'_ny^n}$ with $p_i, p'_i \in K[x,y]$ and $p'_i \notin (x,y)$. But since $x+y$ is not divisible by a power of $y$, we must have $\frac{1}{x+y} = \frac{p_0}{p'_0}$ and therefore $p'_0=x+y$, which is clearly in $(x,y)$, a contradiction.

so $K[x,y]_{(x,y)}[1/y] \subsetneq K[x,y]_{(x)}$

shintuku

02:59

anyhow, we have that it is a subring of a local ring and that's probably local

thanks for the help

Sourav Ghosh

04:51

A bilinear form $B$ on $\Bbb {R}^n$ is an inner product iff matrix of $B$ is positive definite.

Let $A$ is positive definite and $B(X, Y) =Y^tAX $

Then positivity and definiteness of $B$ follows from the positive definiteness of $A$

Linearity in the first slot is also obvious.

Conjugate symmetry of $B$ follows the the symmetry of $A$

gewbDog5

@TedShifrin If I were to watch your lecture series for review

do you have like a curated problem set you would recommend to accompany watching the lectures?

I'm talking about the calc 3/lin. al. series

D Left Adjoint to U

08:37

0

Q: There is a twin prime counting formula that is a direct generalization of the $\pi(x)$ inclusion-exclusion formula found on Wikipedia

I can prove on paper, using elementary techniques that if $f(t) =$ the total number of twin prime averages in the interval $(\sqrt{t+1}, t]$, then we may write: $$ f(t) = \sum_{d \mid p_n \#}(-1)^{\omega(d)} \sum_{r^2 = 1 \pmod d} \lfloor\frac{ t - r}{d} \rfloor $$ for all real numbers $t\in \Bbb...

one potato two potato

10:24

peep

Mad

10:51

$\Pi_i (A - \lambda_i \mathrm{Id}) (\psi) = \Pi_i (A - \lambda_i \mathrm{Id}) \left(\sum_i c_i v_i\right) \Leftrightarrow \Pi_i \left( A(\sum_i c_i v_i) - \lambda_i \mathrm{Id}(\sum_i c_i v_i) \right) = \Pi_i \left( \sum_i c_i (A(v_i) - \lambda_i \mathrm{Id} v_i) \right) = \Pi_i \left( \sum_i c_i (\lambda_i v_i - \lambda_i v_i) \right) = 0
$

is this argumentation correct

i am bit unsure due to the use of index i in both the sum and the product

A id matrices lamda eigenvalues psi is avector vi are the basis vectors and are eigenvectors of A.

technically speaking, i should denjote the sum with j?

Farhad Rouhbakhsh

11:07

@TedShifrin Hello. I have a question regarding my question of a smooth simple closed curve as a union of finite number of arcs, each of them being a smooth function of one variable. I was thinking of an example (although not closed) : Define your curve $C$ by C(t) = (t^3,t^2) for t between -1 and 1. This is infinitely differentiable but has a corner at t=0. At this corner the derivative is (0,0) . So the curve is non-regular.

I observe that by "cutting" the curve at the point (0,0), you can consider the left part and the right part of the curve as a smooth function of y, where the derivative of each function (one variable, regarding to y) is 0 at point x = 0.

Mad

nvm i found my mistake

Koro

12:44

Let Z be the set of all irrational nos. How do I show that $\mathbb R^2\setminus Z^2$ is connected?

one potato two potato

project to one of the axis and connect to the origin (for given point)

Jakobian

13:07

by project potato means to take a path onto the x-axis, parallel to the y-axis, say

depending on which coordinate is rational

Sourav Ghosh

13:21

Let $X, Y$ be two connected spaces and $A, B$ be two proper subsets of $X$ and Y$ respectively. Then $(X\times Y) \setminus (A\times B)$ is connected.

Proof:math.stackexchange.com/a/3395166/977780

Farhad Rouhbakhsh

@TedShifrin I'd be happy to hear your response on my comment. If you want to help, you can do it. Perhaps reading the message with no response saves your time, but is of no use to me... As I told you earlier, I am still stuck to find a non-regular counter example, either by imagining its shape or defining it analytically, and I share my observations with you. I'd be happy to be helped step by step. I am stuck.

shintuku

you have complete freedom to help him ted, you are unshackled by any bounds to help him

we know you thought you couldn't before, so we're teaching you the ways of the world. you are free ted, free!

Farhad Rouhbakhsh

@shintuku hello ?

shintuku

hi

anyhow what's your question

also are you sure the derivative at the point (0,0) is (0,0)

seems to me that it is possible the derivative doesn't exist

Farhad Rouhbakhsh

13:38

@shintuku derivative of (t^3,t^2) at point t = 0 is not (0,0)??

shintuku

well do you have proof

Farhad Rouhbakhsh

Original question
https://math.stackexchange.com/questions/4712503/is-a-smooth-simple-closed-curve-the-union-of-finitely-many-arcs

read the comments if you want to help making a non regular counter example for the result

shintuku

do you have proof the derivative of (t^3, t^2) at (0,0) is (0,0)?

Farhad Rouhbakhsh

@shintuku derivative of a vector consists of the derivative of each of its components

so the derivative of (t^3, t^2) is (3t^2, 2t). clear? Now tell me, and feel free, that what happens when we plug t=0. Feel free to answer.

10th grade math?

Derivative of this function is defined from [0,1] to L(R,R^2). At each point the derivative is a linear transformation, represented by a matrix, and this matrix is usually called the "derivative".

@shintuku No need to interfere when you don't know the basics of the issue.

shintuku

listen you monkey, isn't the derivative of a parametric equation dy/dx

farrhad (bicycle) rubbish

Hopeful Whitepiller

13:50

Sei nett zu ihm

shintuku

hey he started it

Hopeful Whitepiller

Es ist nicht gut, wenn man unhoeflich zu einem anderem Person im Internet ist

Ja, aber wir brauchen kein Kampf hier

Farhad Rouhbakhsh

@shintuku You need to take Advanced calculus or read about it to understand the meaning of multivariable derivative, then we can talk, and of course I am free, like Ted, to answer your insult or not. I prefer not. Best wishes.

Hopeful Whitepiller

this is a good question that Farhad has asked

I had the same one, and I asked on physics stackexchange before

Farhad Rouhbakhsh

@shintuku Yeah, and this is the message where I started, with a post-modern meaning of "I", perhaps an intersubjective "I".

shintuku

13:54

@FarhadRouhbakhsh your curve isn't infinitely differentiable you trash, only the components of the vector are

Hopeful Whitepiller

brudda, wenn du so weitermachst, wirst du ein ban bekommen. Also hoer auf

Farhad Rouhbakhsh

@shintuku I am talking to myself now. Or you are talking to me?

shintuku

"As we discussed, if the parametrization is allowed to be non-regular, the curve may not be smooth at all."

read this sentence from Ted

read it well

read it and understand it

Hopeful Whitepiller

physics.stackexchange.com/questions/703368/… @FarhadRouhbakhsh

shintuku

in other words there's a difference between the derivative of the components of the vector and the differentiability of the curve itself. you need to learn the very very basics of curve parametrization

Farhad Rouhbakhsh

13:58

@shintuku I think it doesn't need insistence from my part to show that I prefer not to talk with you. But you can continue as you like.

Respect is more important than curves

shintuku

okay rubbish, and next time make sure you understand what you're saying before calling someone out on it, go back to 10th math grade (which is indeed your understanding of the subject)

Sourav Ghosh

A smile is a curve that can set a lot of things straight :)

Jakobian

@SouravGhosh broken LaTeX

why assume that $A, B$ are proper? $\emptyset$ is connected

for emphasis maybe

Farhad Rouhbakhsh

14:18

Suppose we have a multivariable function f (here a parametric curve) defined from [0,1] to R^2. The derivative of the curve exists at p if and only if each of its components are differentiable at p. [Pugh, Real Mathematical Analysis, Chapter 5, Theorem 10]. The level of the book is higher than the level of 10th grade math, by the way, so one needs preparation to understand it.

*p is a point in the interval

and (t^3,t^2) is a classical example of a curve that has a cusp, but is nonetheless differentiable at t=0. After reading the book, it may be a good exercise to read some classical examples too, as it helps for beginners.

Sourav Ghosh

@Jakobian Ex:10( Munkres)

I call $\emptyset$ " useless connected set".

Farhad Rouhbakhsh

An interesting case is the curve (a simple circle) defined by (sin(t), cos(t)). When we take the derivative, we see that it is equal to (cos(t), -sin(t)) it is nowhere equal to (0,0). But for another curve with the same image, but a different parametrization, defined by (sin(t^2), cos(t^2)), we have that the derivative is zero at t = 0. So if we change the parametrization it becomes non-regular, If I didn't miss anything.

In the example of (t^3,t^2), the point that the derivative was zero (I mean t = 0) actually identified a cusp in the image. But in the example of (sin(t^2), cos(t^2)), at t=0 we don't have a cusp. As the image of circle is without cusps. I'd be interested if anyone can clarify why this happens, or the intuition behind it.

Jakobian

14:41

@SouravGhosh why useless?

Mad

how can i show that if P is a projection on H_1 then the target set of 1-P is the orthogonal component?
i tried with <v,(1-p)v> to get zero for v in H_1 but no dice

Jakobian

I agree that the empty set as a space is generally useless, but how does connectedness enter the picture

If you're trying to say that connectedness has bad properties because of the empty set, hence it's not really connected, well I disagree with that, empty set is connected, it has bad properties because it's the empty set

The list of properties that work in exception to the empty set is very long, even in set theory

Mad

So i think the statement i madee is only true if pv = v IE P is an orthogonal projection, what would the target set of 1-p be if p is a just a regular projection? the original space?

jay

@porridgemathematics thanks btw that worked !

Mad

can someone help me?

Is it generally true?

i am thinking H\P(H)

because for (1-P)(v)=v-P(v)

Koro

14:58

@Jakobian connected with respect to which def. of connectedness one uses?

Jakobian

15:12

standard definition

Mad

anyone ? any tipp=?

Jakobian

or its equivalent

jay

For a Riemannian manifold with inner product $\langle,\rangle_z$ on the tangent space $T_zZ$. Is the existence of a metric tensor $G(z):T_zZ\to T_zZ^*$ associated to $\langle,\rangle_z$ just Reis representation ?

Sourav Ghosh

15:30

@Mad Your question is not clear.

Empty set is connected [there doesn't exists any non empty proper clopen set].

Thomas Finley

16:13

Hello guys! I was recently putting much thought into the statement: Let F be an Archimedean ordered field. Show that if F
satisfies Cantor's nested interval property then F satisfies Dedekind Completeness property.

I know a proof, that demonstrates an ordered field R (say) [which is a collection of Dedekind cuts (in rational numbers ) ] is complete (i.e every non-empty subset of R has a least upper bound). Can this be considered a proof to the above statement? What are you opinions?

(Also, the proof I have, doesn't even require Cantors Nested Interval Property)

Thorgott

I for one am a proponent of not treating the empty space as connected

Koro

me too

Jaakko Seppälä

16:34

Does anyone know how to prove this: Let $v$ be a discrete valuation of a field $E$, $h\in O_v[X]$ a monic irreducible polynomial of degree $n$, $x$ a root of $h(X)\in\tilde{E}(x)$. Suppose the residue polynomial $\bar h(X) is separable. Then one has to that $v$ is unramified in $F$.

A part of the proof says that the residue map $O_v\to \bar E_v$ can be extended to a place $\varphi_i$ of $F$ with $\varphi_i(x)=a_i$. Here for each $i$ between $1$ and $r$ $a_i$ is a root of $h_i(X)$ in $(\bar E_v)_s$.

Koro

@SouravGhosh added an answer to that just now.

Jakobian

what's the point of expressing an opinion without justifying it

Koro

what?

Jakobian

especially if it's a clashing opinion with someone else's

Koro

I think that it's just push and pull without any backing of an opinion by valid proof.

one party claims their opinion is right, while the other claims themselves to be right. There is no end to this.

Jakobian

16:46

One doesn't have to agree with another, but if you stand by something then you got to have a reason, right?

Otherwise you just blindly agree to something

robjohn

@Jakobian That is commonly called faith.

Jakobian

like believing in the "nature of numbers"?

robjohn

The video? I haven't watched that yet.

Jakobian

In the context of definitions at least, when you are questioning a definition

questioning it without a reason, is just, why

from my standpoint it's irrelevant

monoidaltransform

For $G$ compact Lie group with bi-invariant metric, is it true that the diameter of $G$ bounded above by $\pi?$

Jakobian

16:58

People try to argue that empty set shouldn't be connected because $\prod_{i\in I} X_i$ is connected iff $X_i$ is connected for all $i$ doesn't hold then for example

well, this is the same for pretty much any property that the empty set has, how is that relevant

empty set has weird exceptions like equivalence of $|X|\leq |Y|$ to existence of surjections has an exception with the empty set

or other things connecting with the fact that the empty set has no element, and it's the only set with no element

Sourav Ghosh

Empty set is a "useless finite set".

Jakobian

like, really, any "bad" properties that the empty space has, it just boils down to the empty set being a set with no elements

Koro

it's not useless. You can construct all natural nos. using it.

Jakobian

so I don't see a reason to questions those kind of definitions, rather maybe you want to change foundations

Koro

and then all integers, hence all rationals.

s.harp

17:03

The empty set certainly isn't disconnected, I guess if you are really into topos theory then you could also accept it not being connected

Jakobian

@Koro well really I'd say it depends on your foundation

Koro

continuous maps from emptyset to discrete set {1,2,3}...

Jakobian

I'm pretty confident you can make a version of ZFC where empty set doesn't exist

Koro

does one define functions on an empty set?

Jakobian

adjusting the axioms appropriately

empty set is more like a convenience than a need, probably

but it's what we use in our foundations, so I think it really just depends on what you choose for foundations, in the generally agreed upon ZFC, I'd say it's completely reasonable to say that empty set is connected, or rather, it's unreasonable to say it isn't

Sourav Ghosh

17:08

@Koro Yes.Empty function.

Koro

@冥王Hades Suppose we have a book with cover of one piece or dragon ball characters but inside it is Hatcher's algebraic topology.

Soumik Mukherjee

Just change the cover then

Koro

I'm thinking of doing that actually (putting some OP characters on cover):).

Sourav Ghosh

Suppose $N$ is a normal subgroup of $Gal(L/K)$.Does there always exists a normal extension of $K$ of degree $|N|$?

Ted Shifrin

17:39

@gewbDog5 For problem sets you'll actually need the book. You can find my homework assignments from teaching the course on my old website — these are mostly theoretical problems; the computational problems were done on-line (I wrote literally hundreds of exercises based on ones in the book). alpha.math.uga.edu/~shifrin/MATH3500 and alpha.math.uga.edu/~shifrin/MATH3510

@FarhadRouhbakhsh I was not ignoring you. I have not been here in over 12 hours. Yes, the issue is that you can piece together infinitely many of such curves. Even easier, use joined jagged line segments, but parametrize them so that the derivative is $0$ at each endpoint. In fact, it's even possible to parametrize them so that all derivatives at the endpoints are $0$.

But, as I said, the issue is to put infinitely many of these together (say with corners at $x=1/n$ for all $n$). Picture a shrinking sawtooth function. Anyhow, I guarantee you that Pugh is thinking of a smooth curve as one that has a tangent line at each point; you are on a wild goose chase here.

Thorgott

it's moreso that a connected space should be a space that has exactly $1$ connected component, whereas the empty space has $0$ connected components

if you go to the algebraic geometers, they will unanimously agree that the empty variety/scheme is not irreducible for much the same reason

from a higher topological perspective, connectedness really means $0$-connectedness in an appropriate class of "higher connectivities", each implying the preceeding ones, and being non-empty is the property of being $(-1)$-connected, so should be necessitated by connectedness

i don't really think that just saying "the empty set behaves badly cause it's the empty set" is an argument for not choosing conventions that make a bunch of statements behave as they should

Ted Shifrin

@Thor Well, the empty set is certainly not disconnected. I guess that means it has to be connected, by default. Oh well.

Thorgott

it's the same type of issue as $1$ not being a prime number (even though it's only divisible by $1$ and itself) or the zero ring not being an integral domain (even though all its zero divisors are trivial)

Ted Shifrin

Yeah, I get it. But I just checked Munkres's book and he does not say "nonempty."

Jakobian

I disagree that it's the same issue

Thorgott

17:49

@TedShifrin I certainly wouldn't say it's not disconnected

there's a slight disconnect between intuitive names and convenient formalism going on

Ted Shifrin

Yes you would. You cannot write it as the union of disjoint nonempty open subsets.

Jakobian

anyway, I agree that you can say empty set is not connected when in the context it conceptually helps with things

Thorgott

it defies our intuitive understanding of how the words should work, but I think we should say "connected = not empty + not disconnected", analogous to "irreducible = not empty + not reducible"

Jakobian

It's more the issue between constant polynomial and polynomial of degree 0

where the 0 polynomial sometimes is given degree of -1 or -infinity

Ted Shifrin

Well, I don't find nonempty in the definitions. shrug I don't really care.

Koro

17:53

do we define functions on empty space?

Thorgott

making the empty space connected ruins unique decomposition into connected spaces the same way making 1 a prime ruins unique factorization into primes

@Ted for connectedness? yeah, I don't think anybody really cares, except homotopy theorists. surely you will find it in any AG text for irreducibility, though

shintuku

at koro: emptyset is initial in category Set

Farhad Rouhbakhsh

@TedShifrin Thanks for the response. I thought you were done with my approach or didn't like it :DD. Yeah, the issue is putting together an infinite number of curves with cusps. I've understood until here. I mean I can imagine such curves, by reading your examples. But my problem is I am not sure how to make sure that those curves are smooth, by my definition of smoothness.

Or, in other words, I don't know whether an smooth parametrization exists for the image I have in my mind (for example the image of the sawtooth function). That's where I feel lost. And yeah, I feel I am spending more energy than required, out of curiosity XD. And I feel you are right about the intention of Pugh.

Koro

then $\emptyset$ is connected is same as saying that any continuous function from emptyspace to discrete set {1,2,3} is constant.

Jakobian

sure

constant means there is element of the image such that all elements of domain get mapped to it

no issue there

Thorgott

17:55

yeah, that characterization is potentially inequivalent for the empty space

Ted Shifrin

@FarhadRouhbakhsh You have to define it piecewise. Figure out how to do one line segment, and then define the function by defining it (compatibly) on the intervals $\frac1{n+1}\le x\le \frac 1n$.

Thorgott

if we ask, however, that a space $X$ is connected iff for any discrete space $D$, mapping an element of $D$ to the constant function with that value yields a bijection $C(X,D)\cong D$, then the empty space is not connected

or perhaps we should ask that $C(X,-)$ preserves coproducts, which the empty set equally does not fulfill

not that I think there's an inherent reason to prefer one of these characterizations over another

Koro

@Jakobian but functions are always defined on a non empty domain.

shintuku

at koro: for an arbitrary set s, there is a unique function between empty set and s

Jakobian

are they now

Thorgott

18:00

we should definitely all agree that there is a unique function $\emptyset\rightarrow X$ for any set $X$

this is not negotiable

Koro

then how many functions from $\emptyset$ to $\emptyset$?

shintuku

one

Koro

$0^0$

Jakobian

@Thorgott or decomposition of a set into sets of size $\leq 1$?

Koro

@shintuku so we are accepting that $0^0=1$?

Jakobian

18:01

I think all of those things can be backtracked into cleverly hidden ways that empty set being empty is a culprit, and not really connectedness

shintuku

at koro: for any set $s$, the unique function $\emptyset \to s$ takes an element from $\emptyset$ to $s$. since there is no element to take, any such function is the same

Jakobian

connected set being non-empty for convenience of the theory, I think that is fine

because you won't be saying non-empty over and over again

@Koro If we define $0^0$ to be the number of functions from set of $0$ elements to $0$ elements, then it's not accepting that, it just is that

shintuku

at koro: the reasoning for $\emptyset \to \emptyset$ is the same as with any $\emptyset \to s$

Thorgott

@Jakobian why would you expect a unique decomposition into sets of size $\le1$, though

Jakobian

the issue with $0^0$ can be not $1$ is just that people don't accept it as the number of functions from set of $0$ elements to set of $0$ elements

Koro

18:04

but if range =\emptyset, then where is function $\emptyset\to \emptyset$ mapping to? lol

Mad

i will reformulate my question.

if P is a projection (IE P.P=P and P hermetian)
1-P is also a projection.
if we denote the set that P projects to (or rather the vector space since its linear mapping) as V_1 and the whole vector space as V
where does 1-P (1 is the identity) map to?

I was thinking it was mapping to to the orthogonal elements to V_1

Thorgott

I mean, I'm not here to discuss what the origins of emptiness being an issue are, I'm just here to say that the empty space not being connected is optimal for the theory and I have not seen any argument that would suggest otherwise

Mad

But it appears ( i am not sure) that is only true if P is orthogonal projection

Jakobian

@Thorgott why would you expect unique decomposition into connected spaces

Mad

I have a second question

Thorgott

18:05

cause if we have a non-empty disconnected space, we can split off a component and continue

just like why we would expect unique prime factorization

Ted Shifrin

@Mad Yes. That's called the orthogonal complement of $V_1$.

If $P$ is hermitian, then it is an orthogonal projection.

Jakobian

you can't treat the empty set the same way you treat non-empty set

isn't it just that?

Mad

@TedShifrin but i am not able to prove it, i thought , let v be an element of V_1 , then <v,(1-p)v> is zero, i tried for hours

Jakobian

so it boils down to basic set theory

again!

Ted Shifrin

@Mad That's the wrong inner product.

Mad

18:07

@TedShifrin What do you mean professor?

Ted Shifrin

You want $\langle w,(1-p)v\rangle = 0$ for every $w\in V_1$.

Koro

$\emptyset$ is for stand up comedy.

Mad

where does v come from? V?

shintuku

at koro: it just satisfies the definitions for a function

Thorgott

again, I'm not here to philosophize about why the empty set is different from other sets

Jakobian

18:08

why is connectedness different from any other property

Thorgott

?

shintuku

at koro: e.g. it satisfies $x=y \implies f(x) = f(y)$ because we never have either $x=y$ or $f(x) = f(y)$

Ted Shifrin

@Mad $v$ was yours. Why are you asking me? :D

shintuku

at koro: so it is just a consequence of our definition of a function

Jakobian

Well, whetever or not the reason for why the empty set is different, the empty set is different

Thorgott

18:11

yes, and I think that means its different behavior should be taken into account when making definitions

Jakobian

And like you said, irreducibility for example, we could argue in similar ways for that as for connectedness

shintuku

at koro: if we don't want it to be a function we need to change our definition of functions, or add an ad-hoc element to the definition saying: this is the definition for all functions, except if the relation under consideration is the empty relation. i think this might formally be a contradiction, but without much consequence

Jakobian

So it really boils down to, we assume what's convenient

that's what I think

Mad

Okay, lets say v is from V (whole vectorspace) @TedShifrin then
$<w,v> - <w,Pv> \stackrel{Hermetian}{=} <w,v>-<Pw,v,> $ So it means $<w,pv>=<pw,v> $ this doesnt show anything

Thorgott

and I'm also arguing that for the notion of connectedness, the consistent definition is not make the empty space connected

Jakobian

18:12

I think empty set should be connected unless said otherwise because historically, and in a standard approach, everyone says so

Thorgott

saying "the empty set is different" is not a counter-argument to anything, it's just pointing out why this conversation happens in the first place

Ted Shifrin

Of course not, @Mad. You haven't used either property of $P$. Surely your proof must use both.

Oh, you did use one.

Jakobian

it is, because the argument is that it's not about connectedness

Koro

@shintuku I use (and I think it's the most popular function definition) functions defined on non empty domains.

Thorgott

of course it's about connectedness

Koro

18:13

but there is no reason why this definition can't be extended to emptysets too so there is that.

Thorgott

the empty set warrants exceptions in many definitions, but there's no one-size-fits-all solution

Koro

but emptyset is not connected to me.

shintuku

at koro: i'm thinking about: $f$ function iff (1) $f \subseteq \text{dom} \times \text{im}$ and (2) $x=y \implies f(x) = f(y)$ (3) $f$ is on entire domain

Jakobian

no, it's about the empty set being different because it contains no elements

Thorgott

for any particular definition, you have to consider how to take the empty set into account

Jakobian

18:14

hence doesn't satisfy some properties that all non-empty connected sets satisfy

Mad

well i can then right that the expression is equal due to P^2 = P that
$<pw,pv> $... i still see no zero

Thorgott

yes, you're preaching to the choir, the "why" isn't the point of contention, it is the "what to do with it"

Jakobian

we could replace the property of "non-empty" with a different property like Hausdorff, and try to argue why compact spaces should be Hausdorff

Mad

oh well i guess you can then bring htem on the other side and have zero in teh scalarproduct

Jakobian

the answer is: because it's convenient to assume so

Mad

18:15

Good i mean at this point w could have been equal to v

but now it doesnt make sense, because you could do this proof without 1-p.. jut for p

i am very confused, so p maps to its own orthogonal compliment

In addition, i have noticed
if P and G are projections and they commute, it means by force they are inverse to each other

And my second question was, if P and G anti commute, what does that represent?

Some kind of inverse that maps to -1 instead of 1

Ted Shifrin

@Mad You seem to be writing a very disorganized proof, perhaps assuming what is to be proved.

Mad

@TedShifrin can you guide me on what i can do better?

Farhad Rouhbakhsh

@TedShifrin Good suggestion. I have an example in my mind. For each interval in the x-axis like [1/n, 1/(n+1)], I define my curve as a semi-circle (drawn above the x-axis) with the diameter equal to the length of the interval. The semi-circles keep shrinking up to zero. At the endpoints (where x = 1/n) the curve has cusps, but I think it is also smooth there (not sure about that) because it is where the two tangent semicircles meet . Do you think it works?

Koro

new upvote buttons suck

Ted Shifrin

You have $\langle w,v\rangle - \langle w,Pv\rangle = \langle w,v\rangle - \langle Pw, v\rangle = \langle (w-Pw),v\rangle = 0$ since we assumed $w\in V_1$, so $w=Pw$.

Jakobian

18:23

@Thorgott there is, just remove empty set from your category

Mad

Yes this is what i said

Jakobian

who needs empty topological space?

Ted Shifrin

Definitely NOT smooth when you have a cusp @Farhad. We're still following your bad definition, so you just need to parametrize the zigzag union of infinitely many lines as I suggested, and have a parametrization that has zero derivative at each endpoint.

@Mad It is? I saw you assuming that the equality held and I didn't see you using $w\in V_1$.

Mad

I am sorry i misunderstood. Prof.Ted. why do you equate it to zero? i thought it was what we want to show?

Ted Shifrin

Precisely. I proved it is $0$. I did not just say it was.

Thorgott

18:26

@Jakobian well, good luck intersecting two disjoint subsets of a given space

Ted Shifrin

$w-Pw=0$, and the inner product of $0$ with anything is $0$.

@Thor The intersection is vacuous, hence does not exist.

Mad

But i dont get it, why do you write that equal to zero? how do you know that?

Koro

connected subsets of R with cofinite topo. are finite sets.

Ted Shifrin

Think about what projection means, Mad. What does it mean for $w\in V_1$? What is $Pw$?

shintuku

say i want to argue there is no homomorphism between two rings $R, R'$. is the general approach to this, to show there is no ideal $I$ of $R$ such that $R/I$ has the same number of elements to a subring of $R'$?

Mad

18:27

I understand that Pw = w, i dont understand it mathematically

Ted Shifrin

Well, prove it with the properties.

What is your actual definition of the projection?

Mad

Okay.

P^2= p and P hermetian.
i guess you can <w,pw>=<pw,pw> which follows

ah i see.

Farhad Rouhbakhsh

@TedShifrin OK I will think about your new suggestion. I thought these cusps where the semi-circles meet don't harm smoothness (with my definition of being of a class C infinity), just like the cusp in (t^3,t^2) at t=0 doesn't harm smoothness, and it has zero derivative there.

But it seems I am wrong.

Ted Shifrin

You're confusing your definitions, Farhad. You were working only with parametrizations, not with the actual curves. Now you're switching. And, no, a cusp is NOT a smooth curve.

shintuku

hm i wonder who was saying this

Ted Shifrin

18:30

@Mad The definition I always use is that the projection of $v$ onto $V_1$ is the unique element of $V_1$ with the property that $v-Pv$ is orthogonal to $V_1$.

Mad

well, i am doing quantum mechanics, and the monkeys at my department could care less for understanding what the hell they are mathematically doing, i must Translate everything and fill the gaps, search math books for actually decent rigorous work

My second question was.
suppose P and G are such projections. if PG=GP. it follows PGPG =PPGG=PG so they are by force inverse.

what happens if PG=-GP? we have PGPG = - PPGG = - PG.

Is there any mathematical conclusions to be drawn here?

Ted Shifrin

@Mad Totally wrong.

Mad

@TedShifrin how come?

Ted Shifrin

Nothing to do with inverse.

What if the product is $0$?

Mad

I didnt think of that.

Ted Shifrin

18:34

Or what if $P=G$?

Mad

well thats a trivial case! we do not consider those

Ted Shifrin

Yeah, well, we’ll just ignore the truth.

Mad

No we wont! the question is not formulated decently, i am sorry, the physicst mind is rubbing on me, suppose P and G are different but commute.

I guess it doesnt change the fact that zero is a valuable option

So no inverse..

sad

thank you nevertheless for your assistance

Was very helpful!

Jakobian

@Thorgott But empty set has so many of those properties, that we would have to make pretty much any property to not include the empty set on the same basis. Isn't that tedious? So why change one of them just because of it. Unless you are working ina certain environment for which it would greatly simplify to not use the empty set, I don't see the point

I don't know, I guess this is a boring topic on a boring thing, and it slowly approaches the heights of belief

or well, already reached them

shintuku

is the following correct: the additive group $\mathbb Z_5$ has no subgroup other than the trivial one, since by Lagrange's theorem this would imply the order of the subgroup divides the order of $\mathbb Z_5$, yet 5 is prime

Farhad Rouhbakhsh

18:40

@TedShifrin Oh, I see. But are we not trying to change the shape of the curve such that it can not be written as finitely many arcs? Do you mean that the image of the curve in R^2 with one parametrization is regular, and by the theorem can be split it into finite smooth arcs, and then by changing the parametrization the same can't be done although the image doesn't change? Did I understand correctly or I am deeply confused? XD

Balarka Sen

No neighborhood of the origin in the cusp is a graph over any of the axes, simply because the cusp does not admit a regular parametrization.

Farhad Rouhbakhsh

I mean for example (sin(t),cos(t)) has an image of a circle. Also (sin(t^2),cos(t^2)) has the same image. Your solution works for the first parametrization, because it's regular, but doesn't work for the second, because it has derivate (0,0) at t=0. But nevertheless we can CONSIDER the circle as a union of 2 smooth arcs, either with the first parametrization or with the second. But your solution doesn't work for the 2nd parametrization. That's where I get confused.

shintuku

his answers works, you are hopelessly confused

Ted Shifrin

@Mad actually think about it geometrically instead of just manipulating symbols. What do you conclude about the subspaces?

Mad

i am aware of the geometry, i am actually more interested in rigor of proofs right now.

Balarka Sen

18:46

rigorous proofs don't fall from the sky. if you thought about projections geometrically, it would be obvious to you what commutation of projections meant

Ted Shifrin

@FarhadRouhbakhsh No. That is not at all what we’re doing. You wanted a counterexample without the actual smoothness (regularity).

@Mad I don’t think you are.

Thorgott

@Jakobian no? it's a case-by-case thing

Farhad Rouhbakhsh

@shintuku Read his answer. His answer works only for curves with non-zero derivative. But (sin(t^2),cos(t^2)) has a zero derivative at t = 0. Doesn't it violate the assumption of regularity? But yes, if we change the meaning of "derivative" to having a tangent line at each point, perhaps "your" meaning of derivative, then yes, his answer works because the image remains the same under 2 parametrization.

Thorgott

e.g. I think the empty space should be considered a topological manifold, but I don't think it should be considered connected

shintuku

lol it's not "my" meaning of derivative. don't worry fahrad, that curve isn't smooth, but your brain certainly is

Balarka Sen

18:50

@FarhadRouhbakhsh Let $C$ be image of a smooth map $\gamma : [0, 1] \to \Bbb R^2$ with $\gamma^{(k)}(0) = \gamma^{(k)}(1)$ for all $k \geq 0$. The following are equivalent:
(1) There exists a regular parametrization of $C$, i.e., $\sigma : [0, 1] \to \Bbb R^2$ smooth map with $\sigma^{(k)}(0) = \sigma^{(k)}(1)$ for all $k \geq 0$ such that $\sigma'(t) \neq 0$ for all $t$.
(2) $C \subset \Bbb R^2$ is a *smooth submanifold* ie. for every point $p \in C$ there exists a neighborhood $U \subset \Bbb R^2$ of $p$, and a "tangent line" $\ell$ of $C$ passing through $p$ such that $U \cap C$ is a gra…

Admitting a regular parametrization is, thus, an intrinsic property of the curve $C$.

Farhad Rouhbakhsh

@shintuku Take a look at [Pugh, Chapter 5, Theorem 10]. What I mean by derivative is not dy/dx, I mean the derivative of a function of two variables, with the explanations I told you above. I don't care what you mean by derivative. Regularity means the derivative of all of the components can not be zero at the same time. #Respect

Take a look at this for the definition of regularity (compatible with what I mean the derivative to be): maths.dur.ac.uk/users/pavel.tumarkin/past/fall16/DG/…

shintuku

reread that theorem carefully before citing it, smooth-brain, and stop tagging me

Farhad Rouhbakhsh

@shintuku Take a look at the link. Definition 2.1 (c). I am free to tag or not. #Respect

Ted Shifrin

@Farhad What function of two variables? We don't have any of those. We're not dealing with the curve implicitly.

@shin Just shut up.

shintuku

i will bow to ted's authority and to no one else

also i justify the vitriol because they went on a downvoting spree on me

Jakobian

18:58

@Thorgott isn't that my point? You want to make it case by case but the amount of cases is in abundance

Too tedious, better to just accept the definition on wikipedia or some book

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